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How many integer values ​​of the parameter \(a \in \left( { – 10; + \infty } \right)\) so that the function \(y = \left| {{x^3} + \left( {a + 2} \right)x + 9 – {a^2}} \right|\) covariates over the interval \(\left( {0;1} \right)\)?

A. twelfth.

B. 11.

C. 6.

D. 5.

The answer:

Select REMOVE

Consider \(f\left( x \right) = {x^3} + \left( {a + 2} \right)x + 9 – {a^2}\)

\(f’\left( x \right) = 3{x^2} + a + 2\)

adsense

To \(y = \left| {f\left( x \right)} \right|\) covariates over the interval \(\left( {0;1} \right)\)

TH1:\(\left\{ \begin{array}{l}f’\left( x \right) \ge 0,\forall x \in \left( {0;1} \right)\\f\left( 0 \right) \ge 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}3{x^2} + a + 2 \ge 0,\forall x \in \left( {0;1} \right)\\9 – {a^2} \ge 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a \ge \mathop {Max}\limits_{\left( {0;1} \ right)} \left( { – 3{x^2} – 2} \right)\\9 – {a^2} \ge 0\end{array} \right \Leftrightarrow \left\{ \begin{array }{l}a \ge – 2\\ – 3 \le a \le 3\end{array} \right \Rightarrow a \in \left[ { – 2;3} \right]\)

\(a = \left\{ { – 2; – 1;0;1;2;3;} \right\}\) → 6 values

TH2:\(\left\{ \begin{array}{l}f’\left( x \right) \le ,\forall x \in \left( {0;1} \right)\\f\left( 0 \ right) \le 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}3{x^2} + a + 2 \le 0,\forall x \in \left( {0;1} \right)\\9 – {a^2} \le 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a \le \mathop {Min}\limits_{\left( {0;1} \ right)} \left( { – 3{x^2} – 2} \right)\\9 – {a^2} \le 0\end{array} \right \Leftrightarrow \left\{ \begin{array }{l}a \le – 5\\\left[\begin{array}{l}a\ge3\\a\le–3\end{array}\right\end{array}\right\Rightarrowa\le–5\)[\begin{array}{l}a\ge3\\a\le –3\end{array}\right\end{array}\right\Rightarrowa\le –5\)

Combined with the problem condition \(a = \left\{ { – 9; – 8; – 7; – 6; – 5} \right\}\) → 5 values

So there are 11 satisfying values.